We want to know what degree this curve is.

This was the picture originally shown in La géométrie. Choose an arbitrary straight line KA and fix the ruler (green) at G and place it at an arbitrary point L. 
The same equation will result using any point L. To see a slide show of this, click on the link below. You need ghostview to open this.
First, fix the ruler at the point L. Then construct a line parallel to GA at the point C. This creates the line CB with a length of y. 
Now, connect C and K and construct NL to be parallel to CB and GA. 
Zooming in on triangle CBK, we can see that triangle CBK is similar to triangle NLK since all three angles are equal. Thus, all the sides are proportional. So c/b = y/BK, making BK = yb/c 
To see a proof on similar triangles, click on the link below. This proof is by David Hilbert (18621943).
Again, you need ghostview to open it.
Therefore, BL = yb/c  b 
BK = yb/c BL = yb/c  b AK = x + yb/c  b 
Triangle GAL is similar to triangle CBL since all three angles are equal.

y/(yb/c  b) = a/(x +
yb/c  b) y(x + yb/c  b) = a(yb/c  b) yx + y^{2}b/c  yb = yab/c ab y^{2}b/c = yab/c  ab + yb  yx y^{2} = ya  ac + yc  yxc/b Since the highest degree is two, this curve is a conic section. In fact, it is a hyperbola. 
Now that we know it's a hyperbola, the next problem that we want to address are the asymptotes.
Descartes ended his proof here without addressing this problem.
This next construction and proof is by Van Schooten.
From the slide show, we saw that EA = NL. Thus, EA = c and GA = a. 
Extend AG to make a point D with DG = EA. Then DG = NL = c. 
Draw DF parallel to KC. Claim: DF and AF are the asymptotes to the hyperbola GCE. 
Proof: Extend BC to intersect DF at I. 
Draw DH parallel to AF intersecting BC at H. Triangle NKL is similar to triangle IDH since all three angles are equal. 
Thus, KL/LN = DH/HI DH = AB = x, KL = b, and LN = c. Therefore b/c = x/HI, so HI = cx/b Since HB = DA, HB = c + a Therefore IB = a + c  cx/b and IC = a + c  cx/b  y (IC)(BC) = (DE)(EA), a property of a hyperbola with respect to the distances between it and its asymptotes. Thus, (a + c  cx/b  y)y = ac or y^{2} = cy  cxy/b + ay  ac, which is the same equation as before. Therefore, AF and DF are the asymptotes to the hyperbola GCE. 